函数y=
|
∵函数y=
sin2x+sin2x=1 2
sin2x+1 2
=1-cos2x 2
+1 2
(2 2
sin2x+2 2
cos2x)2 2
=
+1 2
sin(2x-2 2
),∵-1≤sin(2x-π 4
)≤1,∴-π 4
≤2 2
sin(2x-2 2
)≤π 4
,2 2
∴
-1 2
≤y≤2 2
+1 2
,2 2
故函数的值域为 [-
+2 2
,1 2
+2 2
],1 2
故答案为 [-
+2 2
,1 2
+2 2
].1 2